A red die, a blue die, and a yellow die (all six sided) are rolled. we are interested in the probability that the number appearing on the blue die is less than that appearing on the yellow die, which is less than that appearing on the red die. that is, with b, y, and r denoting, respectively, the number appearing on the blue, yellow, and red die, we are interested in p(b < y < r).

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14-01-2018
Mathematics

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A red die, a blue die, and a yellow die (all six sided) are rolled. we are interested in the probability that the number appearing on the blue die is less than that appearing on the yellow die, which is less than that appearing on the red die. that is, with b, y, and r denoting, respectively, the number appearing on the blue, yellow, and red die, we are interested in p(b < y < r).

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W0lf93 W0lf93 · Answer 1 · 2018-01-27T00:35:04+01:00

5/54 or approximately 0.092592593
There are 6^3 = 216 possible outcomes of rolling these 3 dice. Let's count the number of possible rolls that meet the criteria b < y < r, manually.
r = 1 or 2 is obviously impossible. So let's look at r = 3 through 6.
r = 3, y = 2, b = 1 is the only possibility for r=3. So n = 1
r = 4, y = 3, b = {1,2}, so n = 1 + 2 = 3
r = 4, y = 2, b = 1, so n = 3 + 1 = 4
r = 5, y = 4, b = {1,2,3}, so n = 4 + 3 = 7
r = 5, y = 3, b = {1,2}, so n = 7 + 2 = 9
r = 5, y = 2, b = 1, so n = 9 + 1 = 10

And I see a pattern, for the most restrictive r, there is 1 possibility. For the next most restrictive, there's 2+1 = 3 possibilities. Then the next one is 3+2+1
= 6 possibilities. So for r = 6, there should be 4+3+2+1 = 10 possibilities.
Let's see
r = 6, y = 5, b = {4,3,2,1}, so n = 10 + 4 = 14
r = 6, y = 4, b = {3,2,1}, so n = 14 + 3 = 17
r = 6, y = 3, b = {2,1}, so n = 17 + 2 = 19
r = 6, y = 2, b = 1, so n = 19 + 1 = 20
And the pattern holds. So there are 20 possible rolls that meet the desired criteria out of 216 possible rolls. So 20/216 = 5/54.